ce gate set 2 with solutions - ace engineering academy

: 2 : GATE_2016_CE

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Q. 1 – Q. 5 carry one mark each.

01. If I were you, I________ that laptop. It’s much too expensive

(a) won’t buy (b) shan’t buy (c) wouldn’t buy (d) would buy

01. Ans: (c)

Sol: In if clause (type2) ‘were’ is in the past tense so the so main clause should be in the conditional

clause. Therefore ‘C’ is the best answer

02. He turned a deaf ear to my request. What does the underlined phrasal verb mean?

(a) Ignored (b) appreciated (c) twisted (d) returned

02. Ans: (a)

Sol: ‘turned a deaf ear’ means ignored

03. Choose the most appropriate set of words from the options given below to complete the following

sentence.

_____ ______ is a will, ______ is a way.

(a) Wear, there, their (b) Were, their, there (c) Where, there, there (d) Where,their, their

03. Ans: (c)

Sol: Where there is a will there is a way. It is a quotation

04. (x% of y) + (y% of x) is equivalent to _____

(a) 2% of xy (b) 2% of (xy/100) (c) xy% of 100 (d) 100% of xy

04. Ans: (a)

Sol: (x% of y) + (y% of x) = x y

y x100 100

xy xy

100

2xy

100

= 2% of xy

: 3 : SET_2_Forenoon Session


: 4 : GATE_2016_CE


05. The sum of the digits of a two digit number is 12. If the new number formed by reversing the digits

is greater than the original number by 54, find the original number.

(a) 39 (b) 57 (c) 66 (d) 93

05. Ans: (a)

Sol: The new number formed by reversing the digits is greater than the original number is possible in

options A and B only.

Options ‘a’:

Sum of two digits in the number = 3 + 9 = 12

After reversing the two digits number = 93

The difference between the new number formed and original number = 93 – 39 = 54

option ‘A’ is correct.

Option ‘b’:

original number = 57

After reversing, the number is formed = 75

The difference between these two numbers = 75 – 57 = 18

It is not.

06. Two finance companies, P and Q, declared fixed annual rates of interest on the amounts invested

with them. The rates of interest offered by these companies may differ from year to year. Year-wise

annual rates of interest offered by these companies are shown by the line graph provided below.

If the amounts invested in the companies, P and Q, in 2006 are in the ratio 8:9, then the amounts

received after one year as interests from companies P and Q would be in the ratio:

(a) 2:3 (b) 3:4 (c) 6:7 (d) 4:3

2000 2001 2002 2003 2004 2005 2006

6.5 8 8 8

7.5 6.5

4

6

8 9 10 9.5 9

7 P Q



06. Ans: (d)

Sol: The amounts invested in the companies of, P and Q in 2006 = 8 : 9

The rate of interest of company ‘P’ in 2006 = 6%

The rate of interest of company ‘Q’ in 2006 = 4%

The amounts received after one year by P and Q companies in 2006 year

P Q

6% of 8 : 4% of 9

6

8100

: 4

9100

4 : 3

07. Today, we consider Ashoka as a great ruler because of the copious evidence he left behind in the

form of stone carved edicts. Historians tend to correlate greatness of a king at his time with the

availability of evidence today.

Which of the following can be logically inferred from the above sentences?

(a) Emperors who do not leave significant sculpted evidence are completely forgotten.

(b) Ashoka produced stone carved edicts to ensure that later historians will respect him.

(c) Statues of kings are reminder of their greatness.

(d) A king’s greatness, as we know him today, is interpreted by historians.

07. Ans: (d)

Sol: ‘Today, historians correlate greatness of a king at his time with the availability of evidence.’ This

statement leads to the best inference option ‘d’

08. Fact 1: Humans are mammals.

Fact 2: Some humans are engineers.

Fact 3: Engineers build houses

If the above statements are facts, which of the following can be logically inferred?

I. All mammals build houses.

II. Engineers are mammals.

III. Some humans are not engineers.

(a) II only (b) III only (c) I, II and III (d) I only

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08. Ans: (b)

Sol: From given facts, the following venn diagram is possible.

H = Humans

M = Mammals

E = Engineers

BH = Build houses

From above diagram, statement III is true.

09. A square pyramid has a base perimeter x, and the slant height is half of the perimeter. What is the

lateral surface area of the pyramid?

(a) x2 (b) 0.75 x2 (c) 0.50 x2 (d) 0.25 x2

09. Ans: (D) Sol: Base perimeter of square pyramid = x = p

Slant height 2

xl

Lateral surface area of pyramid

2

1 Base perimeter (p) slant height (l)

BH

E

M

H

x/2

x/4 x/4



2

x p

2

1l

2

xx

2

1

4

x2

= 0.25x2

10. Ananth takes 6 hours and Bharath takes 4 hours to read a book. Both started reading copies of the

book at the same time. After how many hours is the number of pages to be read by Ananth, twice

that to be read by Bharath? Assume Ananth and Bharath read all the pages with constant pace.

(a) 1 (b) 2 (c) 3 (d) 4

10. Ans: (c)

Sol: Ananth takes 6 hours to read a book

Bharath takes 4 hours to read a book

L.C.M = 12

The number of pages read by Ananth and Bharath must be 12 (or) multiple of 12 only.

If Ananth read 12 number of pages in 6 hrs

1 hr = 12

6 = 2 pages

If Bharath read 12 number of pages in 4 hrs

1 hr = 12

4 = 3 pages

Option ‘a’

it is not

1 hrs

Ananth Bharath

Read 2 pages

Not read 10 pages

Read 3 pages

Not read 9 pages

10 : 9

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Option ‘b’

it is not

Options ‘c’

After 3 hours is the number of pages to be read by Ananth, twice that to be read by Bharath.

3 hrs

Ananth Bharath

Read 6 pages

not Read 6 pages

Read 9 pages

not Read 3 pages

2 : 1

2 hrs

Ananth Bharath

Read 4 pages

Not read 8 pages

Read 6 pages

Not read 6 pages

8 : 6

4 : 3



Q1 – Q. 25 carry One Mark each

01. The spot speeds (expressed in km/hr) observed at a road section are 66, 62, 45, 79, 32, 51, 56, 60, 53

and 49. The median speed (expressed in km/hr) is_______

(Note: answer with one decimal accuracy)

01. Ans: 54.5

Sol: Increasing order of spot speeds

32, 45, 49, 51, 53, 56, 60, 62, 66, 79

Median = middle values average 2

5653 = 54.5 kmph

02. The optimum value of the function f(x) = x2 4x + 2 is

(a) 2 (maximum) (b) 2 (minimum) (c) 2 (maximum) (d) 2 (minimum)

02. Ans: (d)

Sol: f(x) = x2 – 4x +2

f(x) = 2x – 4 = 0 x = 2

f(x) = 2 > 0 minimum at x = 2

min value is f(2) = 4 – 8 +2 = –2

03. The fourier series of the function,

f(x) = 0, – < x 0

= –x, 0 < x <

In the interval [–,] is

....

3

x3sin

2

x2sin

1

xsin.....

23

x3cos21

xcos2

4xf

The convergence of the above Fourier series at x = 0 gives

(a)

1n

2

2 6n

1 (b)

12n

1 2

1n2

1n

(c)

1n

2

2 81n2

1 (d)

1n

1n

41n2

1

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03. Ans: (c)

Sol: at x = 0

0 =

...

3

1

1

12

4 22

1n21n2

12

4

1n21n2

12

4

1n2

2

1n2

1

8



04. X and Y are two random independent events. It is known that P(X) = 0.40 and P(XYC) = 0.7.

Which one of the following is the value of P(XY) ?

(a) 0.7 (b) 0.5 (c) 0.4 (d) 0.3

04. Ans: (c)

Sol: Given P(X) = 0.4

P(XY ) = 0.7

P(X) + P( Y ) – P(XY ) = 0.7

9

3)Y(P,

9

6)Y(P

P(XY) = P(X)+ P(Y) – P(XY) = 4.05

2

05. What is the value of ?yx

xylim

220x0y

(a) 1 (b) 1 (c) 0 (d) Limit does not exist

05. Ans: (d)

Sol: 220x yx

xylt

It is in the form 0

0

Put y = mx

As y 0, mx 0

222

2

0x m1

m

m1x

mxlt

For different values of m we get different limits it is not unique therefore limit does not exist.

: 12 : GATE_2016_CE


06. The kinematic indeterminacy of the plane truss shown in the figure is

(a) 11 (b) 8 (c) 3 (d) 0

06. Ans: (a)

Sol: Dk = 2j –r ; j = 7; r = 3

Dk = 2 × 7 – 3 = 11

07. As per IS 456-2000 for the design of reinforced concrete beam, the maximum allowable shear stress

(c max) depends on the

(a) grade of concrete and grade of soil

(b) grade of concrete only

(c) grade of steel only

(d) grade of concrete and percentage of reinforcement

07. Ans: (b)

Sol: max depends on grade of concrete as per IS: 456-2000



08. An assembly made of a rigid arm A-B-C hinged at end A and supported by an elastic rope C-D at

end C is shown in the figure. The members may be assumed to be weightless and the lengths of the

respective members are as shown in the figure.

Under the action of a concentrated load P at C as shown, the magnitude of tension developed in the

rope is

(a) 2

P3 (b)

2

P (c)

8

P3 (d) P2

08. Ans: (b)

Sol: MA = 0

P ×l – TX × l – TY × l = 0

2 × T × P2

1

2

P

2

2PT

09. As per Indian standards for bricks, minimum acceptable compressive strength of any class of burnt

clay bricks in dry state is

(a) 10.0 MPa (b) 7.5 MPa (c) 5.0 MPa (d) 3.5 MPa

09. Ans: (d)

Sol: Minimum compressive strength of burnt clay brick = 3.5 MPa

L

CL L

Rigid arm

RopeL

A D

B

P

lB C

45o

TX = Tcos45o

T

TY =T sin45o

Pl

HA

VA

A

: 14 : GATE_2016_CE


10. A construction project consists of twelve activities. The estimated duration (in days) required to

complete each of the activities along with the corresponding network diagram is shown below.

Activity Duration

(days)

Activity Duration

(days)

A Inauguration 1 G Flooring 25

B Foundation work 7 H Electrification 7

C Structural construction-1 30 I Plumbing 7

D Structural construction-2 30 J Wood work 7

E Brick masonry work 25 K Coloring 3

F Plastering 7 L Handing over function 1

Total floats (in days) for the activities 5-7 and 11-12 for the project are, respectively,

(a) 25 and 1 (b) 1 and 1 (c) 0 and 0 (d) 81 and 0

10. Ans: (c)

Sol:

4 6 8

5 7 9

10 11 12 1 2 3

A B C

E F

D G H

I

J

K L

4 6 8

5 7 9

10 11 12 1 2 3 A(1)

C(30)

E(25) F(7)

D)30)

G(25) H(7)I(7)

J(7)

K(3) L(1)

0 0 1 1 8 8

38 38 63 63 70 70

77 77 80 80 81 81

70 7063 6338 38

B(7) (0)



Path Duration

A–B–C–E–F–J–K–L 81

A–B–C–Dummy–G–H–I–K–L 81

A–B–D–G–H–I–K–L 81

All are critical activities, hence total floats of the activities 5–7 and 11-12 are 0 and 0 (default)

11. A strip footing is resting on the surface of a purely clayey soil deposit. If the width of the footing is

doubled, the ultimate bearing capacity of the soil.

(a) becomes double (b) becomes half (c) becomes four-times (d) remains the same

11. Ans: (d)

Sol: For pure clays, the bearing capacity is independent of the footing width.

12. The relationship between the specific gravity of sand (G) and the hydraulic gradient (i) to initiate

quick condition in the sand layer having porosity of 30% is

(a) G = 0.7i+1 (b) G = 1.43 i –1 (c) G = 1.43 i + 1 (d) G = 0.7 i –1

12. Ans: (c)

Sol: At quick condition, i = ic

i = (G 1) (1 n)

= (G 1) (1 0.3)

= ( G 1) (0.7)

G = 1.43 i + 1

13. The results of a consolidation test on an undistributed soil, sampled at a depth of 10 m below the

ground level are as follows:

Saturated unit weight : 16 kN/m3

Pre-consolidation pressure : 90 kPa

The water table was encountered at the ground level. Assuming the unit weight of water as 10

kN/m3, the over-consolidation ratio of the soil is:

(a) 0.67 (b) 1.50 (c) 1.77 (d) 2.00

: 16 : GATE_2016_CE


13. Ans: (b)

Sol: At 10 m depth, = 10

= [16 10] 10 = 60 kPa

OCR = 50.160

90c

14. Profile of a weir on permeable foundation is shown in figure I and an elementary profile of

‘upstream pile only case’ according to Khosla’s theory is shown in figure II. The uplift pressure

heads at key points Q, R and S are 3.14 m, 2.75 m and 0 m, respectively (refer figure II).

What is the uplift pressure head at point P downstream of the weir (junction of floor and pile as

shown in the figure I)?

(a) 2.75 m (b) 1.25 m (c) 0.8 m (d) Data not sufficient

14. Ans: (b)

Sol: hR = h – hP

2.75 = 4 – hP

hP = 4 – 2.75 = 1.25 m

1 m

3 m

10 m 5 m 25 m P 5 m

Gate

Weir

Floor

Figure-I

40 m

R5 m

Q

4 m

S

Figure-II



15. Water table of an aquifer drops by 100 cm over an area of 1000 km2. The porosity and specific

retention of the aquifer material are 25% and 5%, respectively. The amount of water (expressed in

km3) drained out from the area is ______

15. Ans: 0.2

Sol: Sy = n – Sr = 25 – 5 = 20% = 0.2

Amount of water drained = Sy Thickness of aquifer Surface area

= 100

0.2 1000100 1000

= 0.2 Km3

16. Group I contains the types of fluids while Group II contains the shear stress-rate of shear relationship

of different types of fluids, as shown in the figure

Group-I Group-II

P. Newtonian fluid 1. Curve 1

Q. Pseudo plastic fluid 2. Curve 2

R. Plastic fluid 3. Curve 3

S. Dilatant fluid 4. Curve 4

5. Curve 5

The correct match between Group I and Group II is

(a) P-2, Q-4, R-1, S-5 (b) P-2, Q-5, R-4, S-1

(c) P-2, Q-4, R-5, S-3 (d) P-2, Q-1, R-3, S-4

16. Ans: (c)

5

4 2 3

1 Yie

ld s

tres

s sh

ear

stre

ss

Rate of shear

: 18 : GATE_2016_CE


17. The atmospheric layer closest to the earth surface is

(a) the mesosphere (b) the stratosphere (c) the thermosphere (d) the troposphere

17. Ans: (d)

18. A water supply board is responsible for treating 1500 m3/day of water. A settling column analysis

indicates that an overflow rate of 20 m/day will produce satisfactory removal for a depth of 3.1 m. It

is decided to have two circular settling tanks in parallel. The required diameter (expressed in m) of

the settling tanks is_______

18. Ans: 6.9

Sol: Q = 1500 m3/day v0 = 20 m/day

Total surface area required = ov

Q =

1500

20 = 75 m2

Surface area of each setting tank = 75

2 = 37.5 m2

2d4

= 37.5

d = 6.9 m

Diameter of setting tank d = 6.9 m

19. The hardness of a ground water sample was found to be 420 mg/L as CaCO3. A softener containing

ion exchange resins was installed to reduce the total hardness to 75 mg/L as CaCO3 before supplying

to 4 households. Each household gets treated water at a rate of 540 L/day. If the efficiency of the

softener is 100%, the bypass flow rate (expressed in L/day) is_______



19. Ans: 385.714

Sol: Cin = 420 mg/l

Cout = 75 mg/l

Q = 4 540 = 2160 lit/day

= in out

in

C C100

C

=

420 75100

420

= 82.142%

= 100%

Q

CQC)QQ(C BBoutB

mix

2160

420Q0)QQ(75 BB

QB = 385.714 lit/day

Flow that can be by passed = 385.714 Lit/day

20. The sound pressure (expressed in µPa) of the faintest sound that a normal healthy individual can

hear is

(a) 0.2 (b) 2 (c) 20 (d) 55

20. Ans: (C)

Q(Q – QB)

(Q – QB)

Q 75 mg/l 4540

QB

CB= C = 420 = 100%

Cout = 0

Cout = 0

Cin = 420

: 20 : GATE_2016_CE


21. In the context of the IRC 58-2011 guidelines for rigid pavement design, consider the following pair

of statements.

I. Radius of relative stiffness is directly related to modulus of elasticity of concrete and inversely

related to Poisson’s ratio

II. Radius of relative stiffness is directly related to thickness of slab and modulus of subgrade

reaction.

Which one of the following combinations is correct?

(a) I: True; II: True (b) I: False; II: False (c) I: True; II: False (d) I: False; II: True

21. Ans: (b)

Sol: Radius of relative stiffness, 4

1

2

3

)1(k12

Eh

Statement -1: False

Directly proportional to modulus of elasticity and also

( As increases l decreases)

Statement-2: False

22. If the total number of commercial vehicles per day ranges from 3000 to 6000, the minimum

percentage of commercial traffic to be surveyed for axle load is

(a) 15 (b) 20 (c) 25 (d) 30

22. Ans: (a)

23. Optimal flight planning for a photogrammetric survey should be carried out considering

(a) only side- lap (b) only end-lap

(c) either side-lap or end-lap (d) both side-lap as well as end-lap

23. Ans : (d)



24. The reduced bearing of a 10 m long line is N30oE. The departure of the line is

(a) 10.00 m (b) 8.66 m (c) 7.52 m (d) 5.00 m

24. Ans: (d)

Sol: Departure = D = l sin = 10 sin30o = 5.0 m

25. A circular curve of radius R connects two straights with a deflection angle of 60. The tangent length

is

(a) 0.577 R (b) 1.155 R (c) 1.732 R (d) 3.464 R

25. Ans: (a)

Sol: Tangent = T = R577.030tanR2

tanR o

Q.26 – Q.55 carry Two Mark each

26. Consider the following linear system

x+ 2y –3z = a

2x + 3y + 3z = b

5x + 9y – 6z = c

This system is consistent if a,b and c satisfy the equation

(a) 7a – b –c = 0 (b) 3a + b – c = 0 (c) 3a – b + c = 0 (d) 7a – b + c = 0

26. Ans: (b)

Sol:

c695

b332

a321

)AB(

(R2 – 2R1); (R3 – 5R1)

a5c910

a2b910

a321

(R3 – R2)

: 22 : GATE_2016_CE


)a3bc(000

a2b910

a321

(c –b–3a) = 0

3a + b – c = 0

27. If f(x) and g(x) are two probability density functions,

otherwise:0

ax0:1a

x

0xa:1a

x

xf

otherwise:0

ax0:a

x

0xa:a

x

xg

Which of the following statements is true?

(a) Mean of f(x) and g(x) are same; Variance of f(x) and g(x) are same.

(b) Mean of f(x) and g(x) are same; Variance of f(x) and g(x) are different.

(c) Mean of f(x) and g(x) are different; Variance of f(x) and g(x) are same.

(d) Mean of f(x) and g(x) are different; Variance of f(x) and g(x) are different.

27. Ans: (b)

Sol:

dxxxfxfE

=

0

a

a

0

22

dxxa

xdxx

a

x



= a

0

230

a

23

2

x

a3

x

2

x

a3

x

=

2

a

3

a

2

a

3

a 2222

= 0

a

0

20

a

2

dxa

xdx

a

xxgE

= a

0

30

a

3

a3

x

a3

x

= 3

a

3

a 22

= 0

22 XEXExfV

0

a

a

0

23

23

2 dxxa

xdxx

a

xXE

= a

0

340

a

34

3

x

a4

x

3

x

a4

x

=

3

a

4

a

3

a

4

a 3333

6

aXE

32

6

axfV

3

22 XEXExgV

dxa

xdx

a

xXE

a

0

30

a

32

= a

0

40

a

4

a4

x

a4

x

: 24 : GATE_2016_CE


= 4

a

4

a0

33

= 2

a

4

a2 33

2

axgV

3

28. The angle of intersection of the curves

x2 = 4y and y2 = 4x at point (0,0) is

(a) 0o (b) 30o (c) 45o (d) 90o

28. Ans: (d)

Sol: Angle between the curves is angle between the tangents at the point of intersection.

29. The area between the parabola x2 = 8y and the straight line y = 8 is ___________

29. Ans: 85.33

Sol:

Area = 8

0

dyy82

= 33.85

2

3y

82

8

0

2

3

(-8, 8) (8, 8)

x2 = 8y X

Y

y = 8

: 26 : GATE_2016_CE


30. The quadratic approximation of f(x) = x3 – 3x2 –5 at the point x = 0 is

(a) 3x2 – 6x –5 (b) –3x2 –5 (c) –3x2 + 6x – 5 (d) 3x2 – 5

30. Ans: (b)

Sol: Quadratic Approximation

= 0''f!2

0x0'f0x0f

2

50f5x3xxf 23

00'fx6x3x'f 2

f = 6x – 6 f (x) = –6

Equation is = 62

x0x5

2

5x3 2

31. An elastic isotropic body is in a hydrostatic state of stress as shown in the figure. For no change in

the volume to occur, what should be its Poisson’s ratio?

(a) 0.00 (b) 0.25 (c) 0.50 (d) 1.00

31. Ans: (c)

Sol: For hydrostatic state = 0.5

y

x

z



32. For the stress state (in MPa) shown in the figure, the major principal stress is 10 MPa.

The shear stress is

(a) 10.0 MPa (b) 5.0 MPa (c) 2.5 MPa (d) 0.0 MPa

32. Ans: (b)

Sol:

Major principal stress

2xy

2

yxyx1 22

10

2xy

2

2

55

2

55

xy = 5 MPa

33. The portal frame shown in figure is subjected to a uniformly distributed vertical load w (per unit

length)

The bending moment in the beam at the joint ‘Q’ is

(a) Zero (b) hogging24

wL2

(c) hogging12

wL2

(d) sagging8

wL2

5

5

5

5

Q R

S P

L

L/2

w

: 28 : GATE_2016_CE


33. Ans: (a)

Sol: DS = 0 Determinate structure

MQ = 0

MQP = 0

MQ = 0

FBD of member QP

34. Consider the structural system shown in the figure under the action of weight W. All the joints are

hinged. The properties of the members in terms of length (L), area (A) and the modulus of elasticity

(E) are also given in the figure. Let L,A and E be 1 m, 0.05 m2 and 30 × 106 N/m2, respectively, and

W be 100 kN.

Which one of the following sets gives the correct values of the force, stress and change in length of

the horizontal member QR?

(a) Compressive force = 25 kN; stress = 250 kN/m2, Shortening = 0.0118 m

(b) Compressive force = 14.14 kN; stress = 141.4 kN/m2, Extension = 0.0118 m

(c) Compressive force = 100 kN; stress = 1000 kN/m2, Shortening = 0.0417 m

(d) Compressive force = 100 kN; stress = 1000 kN/m2, Extension = 0.0417 m

45o

45o

90oS

A,EA,E

R45o

45o

A,E A,EL

Q

W

90o

2A,E

HP =0 VP

MQP

P

VQ

l/2

Q



34. Ans: (c)

Sol: 45sin

F

45sin

F

90sin

W SRQS

SRQS F2

WF

45sin

F

90sin

F

45sin

F QPQRQS

W45sin

1

2

WFQR

FQR = 100 kN (Comp.); stress = 2m/kN2

2000

205.0

100

2m/kN1000

Shortening 6

3

6

3

1030

101000

103005.02

2110100

AE

2W

= 0.0417 m

35. A haunched (varying depth) reinforced concreted beam is simply supported at both ends, as shown

in the figure. The beam is subjected to a uniformly distributed factored load of intensity 10 kN/m.

The design shear force (expressed in kN) at the section X-X of the beam is _____

effective span = 20 m

400 m

600 m

5m 10 kN/m X

X

45o

FQS

B

45o

90o

Q

FQP

FQP

FQR

Lami’s triangle

90o

45o

45o

45o45o

W

Lami’s triangle

FQS

FSR

FQS FSR

: 30 : GATE_2016_CE


35. Ans: 65

Sol: BM@ section XX = M = 2

wx)x(

2

w 2

5.2510)5(2

2010

= 375 kN-m

SF @ X = 2

20105 10 = 50 kN

(SF)X = )tan(d

MV u

u

=

10

2.0

5.0

37550

Design SF, (SF)x = 50 + 15 = 65 kN

36. A 450 mm long plain concrete prism is subjected to the concentrated vertical loads as shown in the

figure. Cross section of the prism is given as 150 mm × 150 mm. Considering linear stress

distribution across the cross-section the modulus of rupture (expressed in MPa) is _________

36. Ans: 3

Sol:

6

150150

1501025.11

Z

Mf

2

3

cr 3MPa

150 mm 150 mm 150 mm

SP RQ

11.25 kN 11.25 kN

0.2 m

10 m



37. Two bolted plates under tension with alternative arrangement of bolt holes are shown in figure 1

and 2. The hole diameter, pitch, and gauge length are d, p and g respectively.

Which one of the following conditions must be ensured to have higher net tensile capacity of

configuration shown in figure 2 than that shown in figure 1?

(a) gd2p2 (b) gd4p2 (c) gd4p2 (d) gd4p

37. Ans: (c)

Sol: The net tensile capacity of the plate Tdn = 0.9 An fu/γm1

The net tensile capacity of the plate configuration -1, Tdn1 = 0.9 An1 fu/γm1

The net tensile capacity of the plate configuration -1, Tdn2 = 0.9 An2 fu/γm1

For same material and size of the plate, the net tensile capacity of the plate depends on Net

effective sectional area of plate (An).

Net effective sectional area of plate configuration 1 An1 = (B×t-d×t)

Net effective sectional area of plate configuration 2 An2 = (B×t-2×d×t +p2t/4g = B×t- d×t - d×t +p2t/4g)

[Assume width of plate = B, Thickness of plate = t; Diameter of bolt hole = d; staggered pitch

distance = p and Gauge distance = g]

As per Question Tdn2 > Tdn1 => An2 > An1

=> An2 > An1

=> (B×t-d×t -d×t + p2t/4g) > (B×t-d×t)

=> -d×t + p2t/4g > 0

=> p2t/4g > d × t

=> p2 > 4gd

P Pd

Figure 1

P

g PP

Figure 2

: 32 : GATE_2016_CE


38. A fixed-end beam is subjected to a concentrated load (P) as shown in the figure. The beam has two

different segments having different plastic moment capacities (MP, 2Mp) as shown.

The minimum value of load (P) at which the beam would collapse (ultimate load) is

(a) 7.5 Mp /L (b) 5.0 Mp /L (c) 4.5 Mp /L (d) 2.5 Mp /L

38. Ans: (a)

Sol: 13

4

3

2

21

1P1PPP MM2M2M2P

2

M3M43

2P PP

= 5.5 Mp

PM25.8P

3

21

P×1 = 2 Mp + Mp + Mp + Mp

PM53

2P

PM5.7P

2MP

L L

Mp

2L/3

P

1

2MP

2MP

2MP

4l/32l/3 P

MP

Mechanism-I

MP

2MP

MP

4l/32l/3 P

1

MP

Mechanism-II



39. The activity-on-arrow network of activities for a construction project is shown in figure. The

durations (expressed in days) of the activities are mentioned below the arrows.

The critical duration for this construction project is

(a) 13 days (b) 14 days (c) 15 days (d) 16 days

39. Ans: (c)

Sol:

Path Duration

P–Q–T–W–X : 2 + 3 + 5 + 3 + 2 = 15

P–Q–Dummy–U–W–X : 2+3+0+3+3+2 = 13

P–R–Dummy–U–W–X : 2 + 4 + 0 +3+3+2 = 14

P–R–Dummy–V–X : 2 + 4 + 0 +2+2 = 10

P–S–V–X : 2 + 3 + 2 + 2 = 9

Critical path duration = 15 days

10 20 40 50

30 70

80 90

60

T 5

Q 3

P 2

R 4

U3

W3

X2

V2

5 3

10 20 40 50

30 70

80 90

60

T(5)

Q(3)

P(2) R(4)

U(3)

W(3)

X(2)

V(2)S(3)

(0)

(0)

(0)

: 34 : GATE_2016_CE


40. The seepage occurring through an earthen dam is represented by a flownet comprising of 10

equipotential drops and 20 flow channels. The coefficient of permeability of the soil is 3 mm/min

and the head loss is 5 m. The rate of seepage (expressed in cm3/s per m length of the dam) through

the earthen dam is_______.

40. Ans: 500

Sol: Nf = 20, Nd = 10,

hf = 5m

K = 3 mm/min

sec/m601000

3

d

ff N

N.KhQ

10

205

601000

3

= 5 10–4 m3/sec/m

= 500 cm3/sec/m

41. The soil profile at a site consists of a 5 m thick sand layer underlain by a c- soil as shown in figure.

The water table is found 1 m below the ground level. The entire soil mass is retained by a concrete

retaining wall and is in the active state. The back of the wall is smooth and vertical. The total active

earth pressure (expressed in kN/m2) at point A as per Rankine’s theory is _______

1m

4m

3m

A

Sand

sat = 19 kN/m3, w = 9.81 kN/m3,=32o

sat = 18.5 kN/m3, w = 9.81 kN/m3, c= 25 kN/m2 , =24o

bulk=16.5 kN/m3

C– Soil



41. Ans: 69.67

Sol: Taking 2 = 24 (at point A)

422.0sin1

sin1K

2a

At point A, = 1 + 4 + 3

v = 1 16.5 + 4 (19 9.81) + 3(18.5 9.81)

= 79.33 kPa

hKC2Kp wa2vaa 22

781.9422.025233.79422.0

= 69.67 kPa

42. OMC-SP and MDD-SP denote the optimum moisture content and maximum dry density obtained

from standard Proctor compaction test, respectively. OMC-MP and MDD-MP denote the optimum

moisture content and maximum dry density obtained from the modified Proctor compaction test,

respectively. Which one of the following is correct?

(a) OMC –SP < OMC-MP and MDD-SP < MDD-MP

(b) OMC –SP > OMC-MP and MDD-SP < MDD-MP

(c) OMC –SP < OMC-MP and MDD-SP > MDD-MP

(d) OMC –SP > OMC-MP and MDD-SP > MDD-MP

42. Ans: (b)

: 36 : GATE_2016_CE


43. Water flows from P to Q through two soil samples, Soil 1 and Soil 2, having cross sectional area of

80 cm2 as shown in the figure. Over a period of 15 minutes, 200 ml of water was observed to pass

through any cross section. The flow conditions can be assumed to be steady state. If the coefficient

of permeability of Soil 1 is 0.02 mm/s, the coefficient of permeability of Soil 2 (expressed in mm/s)

would be ______

43. Ans: 0.0455

Sol: 1 ml = 1 cm3

sec/cm6015

200Q 3

A = 80 cm2,

h = 600 300 = 300 mm

= 30 cm

L = 150 + 150 = 300 mm

= 30 cm

Q = K. i. A

A.L

hK

6015

200

8030

30K

K = 2.778 103 cm/sec

Where,

K = Equivalent or Average permeability

300 mm

QP

600 mm

Soil 1 Soil 2

150 mm 150 mm



K = 2.778 103 cm/sec

K = 2.778 102 mm/sec

2

2

1

1

21

K

Z

K

ZZZ

K

2.778 102 =

2K

150

02.0

150150150

2K

1507500 = 107. 99 102

K2 = 0.0455 mm/sec

44. A 4m wide strip footing is founded at a depth of 1.5 m below the ground surface in a c- soil as

shown in the figure. The water table is at a depth of 5.5 m below ground surface. The soil properties

are: c = 35 kN/m2, = 28.63o, sat = 19 kN/m3, bulk = 17 kN/m3 and w = 9.81 kN/m3. The values of

bearing capacity factors for different are given below:

Nc Nq N

15o 12.9 4.4 2.5

20o 17.7 7.4 5.0

25o 25.1 12.7 9.7

30o 37.2 22.5 19.7

Using Terzaghi’s bearing capacity equation and a factor of safety Fs = 2.5, the net safe bearing

capacity (expressed in kN/m2) for local shear failure of the soil is _______

5.5 m

4 m

1.5 m

: 38 : GATE_2016_CE


44. Ans: 298.50

Sol: For local shear failure, use Cm and m to compute bearing capacity

2m m/kN33.2335

3

2C.

3

2C

tan tan3

2m

tan 63.28tan3

2m

m = 19.998 say 20

For m = 20, 7.17NC

4.7Nq

5N (from table given)

BN5.01NDNCF

1q qCmns

54175.014.75.1177.1733.235.2

1

qns = 298.50 kN/m2



45. A square plate is suspended vertically from one of its edges using a hinge support as shown in

figure. A water jet of 20 mm diameter having a velocity of 10 m/s strikes the plate at its mid-point,

at an angle of 30o with the vertical. Consider g as 9.81 m/s2 and neglect the self-weight of the plate.

The force F (expressed in N) required to keep the plate in its vertical position is ________

.

45. Ans: 15.71

Sol:

Moment of forces about hinge point O, for equilibrium in vertical position

F 200 = FH 100

2

30sin.AV

2

FF

o2H

o22 30sin)10()02.0(4

1000

= 15.708 N ≃ 15.71 N

46. The ordinates of a one-hour unit hydrograph at sixty minute interval are 0, 3 , 12, 8, 6, 3 and 0 m3/s.

A two-hour storm of 4 cm excess rainfall occurred in the basin from 10 AM. Considering constant

base flow of 20 m3/s, the flow of the river (expressed in m3/s) at 1 PM is _______.

200 mm

Plate

20 mm

30o

Water jet

F

200 mm

Plate

F

100 mm

FH

Hinge

O

F.B.D

: 40 : GATE_2016_CE


46. Ans: 60

Sol: Given 1hr UHG

2 hr SHG ord @ 1 PM = ?

By method of super position 1 hr UHG 2 hr UHG

Time Ist storm

1hr UHG ord

IInd storm

1hr delayed

1hr UHG ord

2hr DRH ord 2hr UHG ord

2hr DRH/2

10 AM 0 –

11 AM 3 0

12 PM 12 3

1 PM 8 12 20 20/2 = 10

2hr UHG ord @ 1 PM = 10 m3/sec

Runoff R (excess rainfall) = 4 cm

2hr DRH ord @ 1 PM = 10 4 = 40 m3/sec

(R = 4 cm)

2hr SHG ord @ 1 PM = 40 + 20 = 60 m3/sec

Flow of river@1 PM = 60 m3/sec

47. A 3 m wide rectangular channel carries a flow of 6 m3/s. The depth of flow at a section P is 0.5 m.

A flat-topped hump is to be placed at the downstream of the section P. Assume negligible energy

loss between section P and hump, and consider g as 9.81 m/s2. The maximum height of the hump

(expressed in m) which will not change the depth of flow at section P is___________

47. Ans: 0.203

b = 3 m

Q = 6 m3/s

yp = 0.5 m

A1 = 3 0.5 = 1.5 m2



5.1

6

A

QV

11 4 m/s

E1 = E2 + (z)Max

81.92

45.0

g2

VyE

221

11

= 1.315 m

3/1

2

23/1

2

2

2c 381.9

6

gb

Qy

= 0.74 m

E2 = Ec = 2

3 0.74 = 1.112 m

1.315 = 1.112 + (Z)max

(Z)max = 1.315 1.112

= 0.203 m

48. A penstock of 1 m diameter and 5 km length is used to supply water from a reservoir to an impulse

turbine. A nozzle of 15 cm diameter is fixed at the end of the penstock. The elevation difference

between the turbine and water level in the reservoir is 500 m. Consider the head loss due to friction

as 5% of the velocity head available at the jet. Assume unit weight of water = 10 kN/m3 and

acceleration due to gravity (g) = 10 m/s2. If the overall efficiency is 80%, power generated

(expressed in kW and rounded to nearest integer) is ___________.

48. Ans: 6714.6

Sol:

Velocity at the jet = gH2V

500102 100 m/s

Velocity head at the jet g2

V2

= 500 m

: 42 : GATE_2016_CE


Head loss due to friction 500100

5 = 25 m

Net head after losses = 500 25 = 475 m

Discharge = 4

AV

(0.15)2 100 = 1.767 m3/s

Power developed = o. . Q. Hnet

= 0.8 10 1.767 475= 6714.6 kW

49. A tracer takes 100 days to travel from Well-1 to Well-2 which are 100 m apart. The elevation of

water surface in Well-2 is 3 m below that in well-1. Assuming porosity equal to 15%, the

coefficient of permeability (expressed in m/day) is

(a) 0.30 (b) 0.45 (c) 1.00 (d) 5.00

49. Ans: (d)

Sol: actual velocity va = time

cetandis=

100

100 = 1 m/day

but va =v

v = va

v = 1 0.15 = 0.15 m/day

but v = Ki ; i = 3

100

K = i

v =

0.15

3 /100 =

0.15 100

3

= 5 m/day



50. A sample of water has been analyzed for common ions and results are presented in the form of a bar

diagram as shown.

The non-carbonate hardness (expressed in mg/L as CaCO3) of the sample is

(a) 40 (b) 165 (c) 195 (d) 205

50. Ans: (a)

Sol: Ca = 2.65 m.eq/lit Mg = 1.45 m.eq/lit

TH = 2.65 50 + 1.45 50 = 205 mg/l as CaCO3

HCO3 = 3.3 m. eq/lit

TA = 3.3 50 = 165 mg/l as CaCO3

TH > TA

CH = TA = 165 mg/l

NCH = TH – CH = 205 – 165 = 40 mg/l as CaCO3

Non carbonacreous hardness = 40 mg/l as CaCO3

51. A noise meter located at a distance of 30 m from a point source recorded 74 dB. The reading at a

distance of 60 m from the point source would be _________

51. Ans: 67.979

Sol: L60 = L30 – 10

6020log

30

L60 = 74 – 10

6020log

30 = 67.979 dB

Noise reading at a distance 60 m = 67.979 dB

0 2.65 4.10 6.35 6.85

K+Na+Mg2+Ca2+

3HCO

24SO Cl–

3.30 3.90 6.750 meq/L

meq/L

: 44 : GATE_2016_CE


52. For a wastewater sample, the three-day biochemical oxygen demand at incubation temperature of

20oC c20,day3 oBOD is estimated as 200 mg/L. Taking the value of the first order BOD reaction rate

constant as 0.22 day–1, the five-day BOD (expressed in mg/L) of the wastewater at incubation

temperature of 20oC c20,day5 oBOD would be _________

52. Ans: 276.15

Sol: ]e1[Ly 3K0

C203

20o

200 = 0.22 30L 1 e L0 = 413.95 mg/l

]e1[Ly 5K0

C205

20o

= 413.95[1–e–0.22 5] = 276.15 mg/l

5 day BOD@ 200C = 276.15 mg/l

53. The critical flow ratios for a three-phase signal are found to be 0.30, 0.25, and 0.25. The total time

lost in the cycle is 10s. Pedestrain crossings at this junction are not significant. The respective

Green times (expressed in seconds and rounded off to the nearest integer) for the three phases are

(a) 34,28 and 28 (b) 40, 25 and 25 (c) 40, 30 and 30 (d) 50, 25 and 25

53. Ans: (a)

Sol: y = y1 + y2 + y3

= 0.3 + 0.25 + 0.25 = 0.8

Lost time, L = 10 sec

Cycle time, 8.01

5105.1

y1

5L5.1Co

= 100s

Green times:

)LC(y

yG o

11

3475.33)10100(8.0

3.0

28125.28)10100(8.0

25.0G2

G3 = G2 = 28



54. A motorist traveling at 100 km/h on a highway needs to take the next exit, which has a speed limit

of 50 km/h. The section of the roadway before the ramp entry has a downgrade of 3% and

coefficient of friction (f) is 0.35. In order to enter the ramp at the maximum allowable speed limit

the braking distance (expressed in m) from the exit ramp is________

54. Ans: 92.14

Sol: Downward gradient, N = 3%

f = 0.35

)Nf(g2

)v()v(S

2f

2i

b

But 18

5100vi 27.77 m/s

18

550vf = 13.88 m/s

100

335.081.92

88.1377.27S

22

b

Sb = 92.14 m

55. A tall tower was photographed from an elevation of 700 m above the datum. The radial distance of

the top and bottom of the tower from the principal points are 112.50 mm and 82.40 mm,

respectively. If the bottom of the tower is at an elevation 250 m above the datum, then the height

(expressed in m) of the tower is __________

55. Ans: 120.4 m

Sol: Relief Displacement = d = r2 – ro = 112.50 – 82.40

= 30.10 m

50.112

)250700(1.30

r

)hH(dh

2

12

= 120.40 m

ce gate set 2 with solutions - ace engineering academy

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